A 220V, 50 Hz ac source is connected to a series combination of a 3.16 micro farad capacitor , 0.63 H inductor and 600 ohm resistor.
Phase difference between voltage and current :

Answer

To solve the problem of finding the phase difference between voltage and current in a series RLC circuit connected to an AC source, we can follow these steps:

### Step 1: Identify Given Values
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Capacitance (C) = 3.16 µF = 3.16 × 10^(-6) F
- Inductance (L) = 0.63 H
- Resistance (R) = 600 Ω

### Step 2: Calculate Angular Frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]

### Step 3: Calculate Inductive Reactance (X_L)
Inductive reactance (X_L) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = (100 \pi) \times 0.63 = 63 \pi \, \Omega \]

### Step 4: Calculate Capacitive Reactance (X_C)
Capacitive reactance (X_C) is calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100 \pi \times 3.16 \times 10^{-6}} = \frac{1}{0.000316 \pi} \approx \frac{3164.55}{\pi} \, \Omega \]

### Step 5: Calculate the Phase Difference (φ)
The phase difference (φ) between voltage and current in an RLC circuit is given by:
\[ \tan(\phi) = \frac{X_L - X_C}{R} \]
Substituting the values of X_L, X_C, and R:
\[ \tan(\phi) = \frac{63 \pi - \frac{3164.55}{\pi}}{600} \]

### Step 6: Simplify the Expression
Calculating the numerator:
\[ \tan(\phi) = \frac{63 \pi - 1007.85}{600} \]
Calculating the value of \( \tan(\phi) \):
\[ \tan(\phi) \approx \frac{197.35}{600} \approx -1.3489 \]

### Step 7: Calculate the Phase Difference (φ)
Now, we find φ using the arctangent function:
\[ \phi = \tan^{-1}(-1.3489) \approx -0.93 \, \text{degrees} \]

### Final Answer
The phase difference between voltage and current is approximately:
\[ \phi \approx -0.93 \, \text{degrees} \]

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